Proof statement for binary relation or show it is wrong by counterexample

Question

Proof or show following statement is false

$$ (R \circ S) \cap (R \circ T) \subseteq R \circ (S \cap T) $$

Statement is false by counterexample

Let $ \langle x,y \rangle \in (R \circ S) \cap (R \circ T) $

$$ \iff (\langle x, y \rangle \in R \circ S) \wedge (\langle x ,y \rangle \in R \circ T) $$

$$ \iff (\exists w: \langle w,y \rangle \in S) \wedge (\exists z : \langle z, y \rangle \in T) $$

$$ \iff \{ \langle x, w \rangle, \langle x, z \rangle \} \in R $$

$$ \iff \{ \langle w, y \rangle \} \cap \{ \langle z , y \rangle \} = S \cap T = \emptyset $$

$$ \iff \langle x , y \rangle \not\in R \circ (S \cap T) $$

$$ \iff (R \circ S) \cap (R \circ T) \not\subseteq R \circ (S \cap T) $$

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Is this valid?

Answer 1

You seem to be somehow on a right track, but what you write is formally awfully wrong. For example, the staements $$ (\exists w: \langle w,y \rangle \in S) \wedge (\exists z : \langle z, y \rangle \in T)$$ and $$\{ \langle x, w \rangle, \langle x, z \rangle \} \in R $$ are certainly not equivalent (the first says nothing about $R$, the second says nothing about $S,T$), hence should not be connected with an "$\iff$".

However, in the line where you end up writing $S\cap T=\emptyset$, you apparently have the right idea, namely to have $S=\{\langle w,y\rangle\}$ and $T=\{\langle z,y\rangle\}$ ... somehow. Turn this into a concrete counterexample with all elements being different as far as required by the idea:

Let $S=\{\langle 1,3\rangle\}$, $T=\{\langle 2,3\rangle\}$ and $R=\{\langle 0,1\rangle,\langle 0,2\rangle\}$. Then $$ R\circ S\cap R\circ T=\{\langle0,3\rangle\}\cap\{\langle0,3\rangle\}=\{\langle0,3\rangle\}$$ whereas $$ R\circ(S\cap T)=R\circ \emptyset=\emptyset.$$ Hence this $R,S,T$ make a counterexample to the given statement.

or somewhat "minimized":

Let $S=\{\langle 0,1\rangle\}$, $T=\{\langle 1,1\rangle\}$ and $R=\{\langle 0,0\rangle,\langle 0,1\rangle\}$. Then $$ R\circ S\cap R\circ T=\{\langle0,1\rangle\}\cap\{\langle0,1\rangle\}=\{\langle0,1\rangle\}$$ whereas $$ R\circ(S\cap T)=R\circ \emptyset=\emptyset.$$